## Precalculus (6th Edition)

$\color{blue}{\left\{1-2i, 1+2i\right\}}$
Subtract $2x-5$ to both sides of the equation to obtain: $x^2-(2x-5)=2x-5-(2x-5) \\x^2-2x-(-5)=0 \\x^2-2x+5=0$ RECALL: The quadratic equation $ax^2+bx+c=0$ can be solved using the quadratic formula: $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ The given equation has: $a=1, b=-2 c=5$ Substitute these values into the quadratic formula to obtain: $x=\dfrac{-(-2) \pm \sqrt{(-2)^2-4(1)(5)}}{2(1)} \\x=\dfrac{2\pm\sqrt{4-20}}{2} \\x=\dfrac{2\pm\sqrt{-16}}{2} \\x=\dfrac{2\pm\sqrt{16(-1)}}{2} \\x=\dfrac{2\pm 4i}{2} \\x=1 \pm 2i$ Thus, the solution set is $\color{blue}{\left\{1-2i, 1+2i\right\}}$.